Hexagony, 6 bytes

x>i;s@

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Inspired by boboquack's 9-byte solution. The following quote was a big hint:

I'm slightly disappointed that I couldn't get six to work (that would be pushing it with three characters, one output, one redirection and one termination).

How it works (or, How I got to this solution)

A 6-byte program is laid out on a hexagon of side length 2, and the 7th instruction is necessarily a no-op:

 ? ?
? ? ?
 ? .

Since I needed at least one mirror, I tried out various mirrors placed at various places, until I found this:

 A >
B C D
 E .

Assuming the current memory is always positive (and there are no branches/mirrors/IP changers among ABCDE), the IP follows the path

A>ADCB>.CAD>DAC.BE

The cell C is run exactly three times, and the cells right before C are D, B, and A respectively. And the cell E is first run after three runs of C. This is exactly what we want: write one of s, i, or x on the memory and print it, and then terminate!

Now back to the actual source code:

 x >
i ; s
 @ .

And the execution path, linearized, with significant instructions emphasized:

x>xs;i>.;xs>sx;.i@
 ^^^ ^ ^^ ^
 ^^ Print 's'
 ^ ^ Print 'i'
 ^^ Print 'x'
 ^ Terminate

Hexagony, 6 bytes, 17 cycles

s;i>x@

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Now the letters six are in more natural order, and the program finishes one cycle faster.

How it works

For more thorough explanation, see the previous answer below.

 A B
C > D
 E .
Execution path: ABC>CBAD>ADB.>.CE
 ^^ ^^ ^^ ^
 s; i; x; @

Hexagony , 6 bytes, 18 cycles

x>i;s@

Try it online!

Inspired by boboquack's 9-byte solution. The following quote was a big hint:

I'm slightly disappointed that I couldn't get six to work (that would be pushing it with three characters, one output, one redirection and one termination).

How it works (or, How I got to this solution)

A 6-byte program is laid out on a hexagon of side length 2, and the 7th instruction is necessarily a no-op:

 ? ?
? ? ?
 ? .

Since I needed at least one mirror, I tried out various mirrors placed at various places, until I found this:

 A >
B C D
 E .

Assuming the current memory is always positive (and there are no branches/mirrors/IP changers among ABCDE), the IP follows the path

A>ADCB>.CAD>DAC.BE

The cell C is run exactly three times, and the cells right before C are D, B, and A respectively. And the cell E is first run after three runs of C. This is exactly what we want: write one of s, i, or x on the memory and print it, and then terminate!

Now back to the actual source code:

 x >
i ; s
 @ .

And the execution path, linearized, with significant instructions emphasized:

x>xs;i>.;xs>sx;.i@
 ^^^ ^ ^^ ^
 ^^ Print 's'
 ^ ^ Print 'i'
 ^^ Print 'x'
 ^ Terminate